Let's see who's smart here

Let's see who's smart here.

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en.wikipedia.org/wiki/Conditional_probability
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50% fucktard

1/3 faggots. Lrn2statistics

it's 1 in 3

Correct

Nope

easy as fuck

1 in 3 statistically, But realistically you have to take into the account the variables such as mass of both sides of the coin, the way it is flipped, and so on.

Given the caveat of "at least" one, this makes two of three possible combinations effectively equivalent. Therefore it truly is one in two, or 50%.

Obviously, hypothetically, with as many 'flips' on an even coin as possible, ideally approaching a horizontal asymptote to give the most accurate number possible.

25%

nope

the first heads is "given". That means it's definite, and you practically didn't flip it at all. It doesn't mean anything.

Nope

You're still flipping two coins, and measuring the flips where AT LEAST one is heads. Read the question.

oh
wait
i get it
nvm

We read it and you're in illiterare fucking retard. 'Given that at least one will be heads, what is the probability that they will both be heads'... The answer is 50%. You fail.

What about the option of it landing sideways?

I fucking love these threads

That looks like the most difficult way possible to solve this using a computer.

Let me try to explain,
There are four possible outcomes to flipping two coin: (T = tail, H= head)
TT
TH
HT
HH

All these situations have 25% chance to happen.

Now given the fact that at least 1 coin landed on heads, it leaves us with 3 possibe options, not 4:
HH
HT
TH

Because all 3 of these outcomes have the same probability of 33%, the probability of getting HH is 33%.

this

100 percent. Think about it.

50/50 if one coin lands on heads

So by math, that means two coins has a 100/100 chance of being heads (because it's 2 coins we add the 50 percent probability). So no matter what it will be heads.

You fuckers need to go back to school

>variables such as mass of both sides of the coin
hmm
doubtful on this one

No, dope. You don't have to calculate the probability of anything except for the coin that is not determined. And there's a 50% chance that it's going to be heads. The end. It does not matter which one of the coins is heads, because the choice is arbitrary. You can take that out of the equation.

If they are REGULAR coins, there's no way to guarantee that one will land heads.
So, it's 25%

en.wikipedia.org/wiki/Conditional_probability
have fun

33%

...

The answer is 1/3.

As the attached image shows, we have 3 possible outcomes, which are filled in red and green, and a 4th outcome crossed out in gray that we know won't happen because we've already assumed at least one coin will be heads.

Of the 3 remaining possible outcomes, only one is a heads+heads result. Therefore, the chances of getting both coins heads assuming one will definitely be heads, is 1/3, or 33.33333... percent.

basically this but picture form

is there a 1% chance the coin turns out to be a hot trap or something?

The way the question is worded is dumb. It makes it sound like "both tails" can magically never happen. That's why people say 1/2.

.5 heads on 1st flip x .5 heads on 2nd flip = .25 both heads on 1st flip.

25% chance both land on heads on 1st flip

you're literally retarded

It's a given that one coin is heads, or will be heads. So, then it's 50% that the other one will be heads.

190.7 %.

Fitty/fitty

every autist in this thread is forgetting one detail ... it doesn't matter which coin lands on heads first. that's why 33% is wrong.

There are not four possible original combinations, only three. HH, HT, TT. Why? Because the order doesn't matter. HT = TH.

Fucking autists. So with one coin already determined to be H, that leaves exactly two possible combinations: HH and TH.

Put away your calculator and take a look at the real world.

bad bait but I replied anyway

not an argument.

>you're in illiterare fucking

Righto m80

Go physically flip two coins 50 times, record each result. Then tell me how many of those 50 dual-coin flips give you at least one coin as heads.

I'll wait.

...

The correct way to test it IRL would be to put one coin on the table heads-up, then flip the other coin a bunch of times.

>the best way to test see outcome of two coins being flipped is to flip one coin

>given that one of them landed on heads

33.33 (repeating, of course) chance

Well, its a lot better than we usually do.

>either coin

Read the first 5 words of the question.

TWO REGULAR COINS were flipped. This is the trap - read the question.

this means that you exclude any results that don't include at least one heads you fucking moron
that isn't the same as putting one coin heads up and flipping the other
please seek basic education

>the trap

That's really a dumb question.
Why would one of the coins be guaranteed to land heads? No practical use whatsoever for this kind of bullshit except maybe some shitty RPG crit calculation fuck

one coin doesn't have to be guaranteed to land heads for this to work, you can test it irl by just excluding results that don't include a heads.
and it's made to test the basic mathematical competence of a subject

More information is needed to make sure we have the correct answer, but the answer to the question is 1/3, but the question is pretty misleading and i believe that has been done on purpose.

One coin state is independent of the other. So if you tell us that one is on heads it all comes down to the probability of the other coin landing on heads which is 50%

It's not 'guaranteed' - it's asking whats the probability both are heads if at least one lands heads. At first glance it would seem to be 50%, but if you actually flip both coins a lot and measure it, you'll see the answer is actually 1/3.

/thread

you're fundamentally misunderstanding the question

just flipped it a bunch of times and it was always heads
therefore it's actually 100%

Possible coin states after flipping two coins are:

HH
TT
HT/TH

1/3 is two heads.

Kek

We have thises outcomes:
coin1 = head, coin2 = tails
coin1 = tails, coin2 = head
coin1 = head, coin2 = head

Out of these three outcomes one is "both heads", so 1 in 3.

this is on the same level of bs as the riddle 2 coins equal 15 cents and one isn't a nickel, the other one is

Dumbarse detected

Isn't the second part of the question essentially irrelevant?

Oh never mind I see it now.

No, it's not. If it was not there, it will be 1 in 4. With this condition, you ignore the situation when both end up tails and the answer is 1 in 3.

Yall niggas posting in a slide thread

The idea is to ignore the results of any test that do not meet the minimum standard of 'at least one of them landed heads.'

Thankyou.


Being intentionally vague just shows how deep you need to scrape that barrel for some acknowledgement, OP

>1 in 3
Okay, so one coin is confirmed as heads.
If it's 1 in 3, the second coin will come up
>heads
>tails
>______
What's the third option?

You ignore that two tails is possible and you also must assume that one will definitely be heads, which one doesn't matter, so even though you're flipping two coins only one is counted, which one? You'll never know because the question is fkn retarded like yourself and OP

You're being trolled stfu

...or the one that lands on tails...

...

>What's the third option?
spelunking
(look it up)

50%

either they both landed heads, or they didn't.

however there are 1:3 odds against them both landing heads.

1/4, 1/3, and 1/2 are all valid answers depending on how you interpret the question. It's just very poorly defined enough that you could interpret in a way that either of the three answers would be valid.

Do euros even have a heads side to land on? I don't think so but I'm American. Assuming they don't, the probability is 0% that one would land on heads.

its 50%, why would one coins result have any effect on the other?

50/50, its like black jack, if RED has come up the last 10 times then the probability of BLACK isn't increased, its still 50/50

SO, according to the question one of the coins has landed on heads. This means we can instantly remove the situation where we get two tails. That leaves us with 3 combinations remaining, HH HT or TH. Getting HH is one in three given that you receive at least one head.

Stop trying to revive the thread you're a fucking idiot this is is searchable on the internet stop it you pasta fag you're not proving anything kys yourself

33.33%

Maybe I missed it, but I haven't seen the correct answer yet. It's both 33% AND 50%. The question says the information is 'given', but how is it given?

if you're blindfolded, and flip two coins and I tell you, "at least one of the coins is heads". In this case the order doesn't matter so there's a 50% chance the other one is heads.

On the other hand, if you mean what are the odds of getting two heads, discounting the times you get two tails, then the 'HH HT TH TT' people are correct, it's 33%.

IMO all the other answers are variations on theses. The lesson here, be careful of words like 'given'. Always ask 'given from where'.

33.3(3)%
since the initial change is 50% and that's already been wasted you now get 33.3(3)% chance.

>69
This guy gets it. :^)

There's another probability that isn't explained in the task.
This "magic" force which picks with coin will certainly be heads.
If this magic force is also 50-50 then that's an additional probability you have to think of and the true answer is 1/3.
However, there is nothing in the problem that states that this is the case.
Since that isn't explicitly stated the problem is flawed.
For all we know the magic force could have a 90-10 bias in which case everything changes.
Just because it says "either" doesn't mean we should assume a 50-50 split.

You could have proven it mathematiclaay fucking brainlet, why do a shitty experiment, when you can prove it mathematically.

This. It either does or it doesn't.

This.

The TT is a lie